Average Error: 0.6 → 0.9
Time: 6.1s
Precision: binary64
\[\frac{e^{a}}{e^{a} + e^{b}} \]
↓
\[\begin{array}{l} \mathbf{if}\;e^{a} \leq 0.9997404070023854:\\ \;\;\;\;\frac{e^{a}}{1 + \left(b + \mathsf{fma}\left(0.5, b \cdot b, \mathsf{fma}\left(0.16666666666666666, {b}^{3}, e^{a}\right)\right)\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{1 + e^{b}}\\ \end{array} \]
\frac{e^{a}}{e^{a} + e^{b}}
↓
\begin{array}{l}
\mathbf{if}\;e^{a} \leq 0.9997404070023854:\\
\;\;\;\;\frac{e^{a}}{1 + \left(b + \mathsf{fma}\left(0.5, b \cdot b, \mathsf{fma}\left(0.16666666666666666, {b}^{3}, e^{a}\right)\right)\right)}\\

\mathbf{else}:\\
\;\;\;\;\frac{1}{1 + e^{b}}\\


\end{array}

(FPCore (a b) :precision binary64 (/ (exp a) (+ (exp a) (exp b))))
↓
(FPCore (a b)
 :precision binary64
 (if (<= (exp a) 0.9997404070023854)
   (/
    (exp a)
    (+
     1.0
     (+ b (fma 0.5 (* b b) (fma 0.16666666666666666 (pow b 3.0) (exp a))))))
   (/ 1.0 (+ 1.0 (exp b)))))
double code(double a, double b) {
	return exp(a) / (exp(a) + exp(b));
}
↓
double code(double a, double b) {
	double tmp;
	if (exp(a) <= 0.9997404070023854) {
		tmp = exp(a) / (1.0 + (b + fma(0.5, (b * b), fma(0.16666666666666666, pow(b, 3.0), exp(a)))));
	} else {
		tmp = 1.0 / (1.0 + exp(b));
	}
	return tmp;
}

Error

Bits error versus a

Bits error versus b

Target

Original0.6
Target0.0
Herbie0.9
\[\frac{1}{1 + e^{b - a}} \]

Derivation

  1. Split input into 2 regimes

  2. if (exp.f64 a) < 0.99974040700238542

    1. Initial program 0.9

      \[\frac{e^{a}}{e^{a} + e^{b}} \]

    2. Taylor expanded around 0 0.9

      \[\leadsto \frac{e^{a}}{\color{blue}{1 + \left(b + \left(0.5 \cdot {b}^{2} + \left(e^{a} + 0.16666666666666666 \cdot {b}^{3}\right)\right)\right)}} \]

    3. Simplified0.9

      \[\leadsto \frac{e^{a}}{\color{blue}{1 + \left(b + \mathsf{fma}\left(0.5, b \cdot b, \mathsf{fma}\left(0.16666666666666666, {b}^{3}, e^{a}\right)\right)\right)}} \]

    if 0.99974040700238542 < (exp.f64 a)

    1. Initial program 0.6

      \[\frac{e^{a}}{e^{a} + e^{b}} \]

    2. Taylor expanded around 0 0.4

      \[\leadsto \color{blue}{\left(\frac{1}{1 + e^{b}} + \left(0.5 \cdot \frac{{a}^{2}}{1 + e^{b}} + \left(\frac{a}{1 + e^{b}} + \frac{{a}^{2}}{{\left(1 + e^{b}\right)}^{2} \cdot \left(e^{b} + 1\right)}\right)\right)\right) - \left(1.5 \cdot \frac{{a}^{2}}{\left(1 + e^{b}\right) \cdot \left(e^{b} + 1\right)} + \frac{a}{\left(1 + e^{b}\right) \cdot \left(e^{b} + 1\right)}\right)} \]

    3. Simplified0.4

      \[\leadsto \color{blue}{\left(\frac{1}{1 + e^{b}} + \left(\frac{a \cdot a}{{\left(1 + e^{b}\right)}^{3}} + \mathsf{fma}\left(0.5, \frac{a}{1 + e^{b}} \cdot a, \frac{a}{1 + e^{b}}\right)\right)\right) - \mathsf{fma}\left(1.5, \frac{a}{{\left(1 + e^{b}\right)}^{2}} \cdot a, \frac{a}{{\left(1 + e^{b}\right)}^{2}}\right)} \]

    4. Taylor expanded around 0 0.9

      \[\leadsto \color{blue}{\frac{1}{e^{b} + 1}} \]
  3. Recombined 2 regimes into one program.

  4. Final simplification0.9

    \[\leadsto \begin{array}{l} \mathbf{if}\;e^{a} \leq 0.9997404070023854:\\ \;\;\;\;\frac{e^{a}}{1 + \left(b + \mathsf{fma}\left(0.5, b \cdot b, \mathsf{fma}\left(0.16666666666666666, {b}^{3}, e^{a}\right)\right)\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{1 + e^{b}}\\ \end{array} \]

Reproduce

herbie shell --seed 2021211 
(FPCore (a b)
  :name "Quotient of sum of exps"
  :precision binary64

  :herbie-target
  (/ 1.0 (+ 1.0 (exp (- b a))))

  (/ (exp a) (+ (exp a) (exp b))))